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Magazine Chapter 1: Problem 33
Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\sqrt{x+1} $$
Short Answer
Expert verified
The inverse is \( f^{-1}(x) = x^2 - 1 \). Both compositions verify the inverse.
Step by step solution
01
Define the Problem
The given function is \( f(x) = \sqrt{x + 1} \). Our task is to find its inverse, denoted \( f^{-1}(x) \), and then verify the inverse by showing that \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \).
02
Express the Function in Terms of y
Start by setting \( y = f(x) \), which gives us \( y = \sqrt{x + 1} \). Now express \( x \) in terms of \( y \).
03
Solve for x
To find \( x \) in terms of \( y \), square both sides of the equation from the previous step: \( y^2 = x + 1 \). Then solve for \( x \) by subtracting 1 from both sides: \( x = y^2 - 1 \). This expression gives us the inverse function, \( f^{-1}(y) = y^2 - 1 \).
04
Replace y with x for Inverse Function
To express the inverse function in terms of \( x \), replace \( y \) with \( x \). Thus, the inverse function is \( f^{-1}(x) = x^2 - 1 \).
05
Verify Inverse with Composition
First verify that \( f^{-1}(f(x)) = x \). Substitute \( f(x) = \sqrt{x + 1} \) into \( f^{-1} \): \( f^{-1}(f(x)) = (\sqrt{x + 1})^2 - 1 = x + 1 - 1 = x \). This simplifies to confirm the identity.
06
Verify Inverse with Reverse Composition
Now verify that \( f(f^{-1}(x)) = x \). Substitute \( f^{-1}(x) = x^2 - 1 \) into \( f \): \( f(f^{-1}(x)) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = x \). This simplifies to confirm the identity, assuming \( x \geq 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Composition of Functions
The composition of functions is like preparing a multi-step recipe where the output of one function becomes the input for another. For a given function \( f(x) \) and its inverse function \( f^{-1}(x) \), the goal is to see how they interact. Here's what typically goes on in the composition of functions:
- Applying the Inverse: When you plug \( f(x) \) into \( f^{-1}(x) \), you get back to the same initial value \( x \). Mathematically, this is represented as \( f^{-1}(f(x)) = x \).
- Double Checking: If you reverse the process by putting \( f^{-1}(x) \) back into \( f(x) \), it should also simplify to \( x \). This is noted as \( f(f^{-1}(x)) = x \).
Verifying Inverse Functions
Verifying that a function has truly been inverted involves a bit of a dance with equations. After identifying the inverse, it's important to ensure that the original and inverse functions truly undo each other.The steps generally include:
- First Check: Take the composition \( f^{-1}(f(x)) \) and see if it simplifies to \( x \). With \( f(x) = \sqrt{x + 1} \) and \( f^{-1}(x) = x^2 - 1 \), substituting gives \( f^{-1}(\sqrt{x + 1}) = (\sqrt{x + 1})^2 - 1 = x \).
- Second Check: Do the reverse composition, \( f(f^{-1}(x)) \), and check that it also simplifies to \( x \). Using the inverses we found, this gives \( f(x^2 - 1) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = x \), assuming \( x \geq 0 \).
Square Root Function
The square root function, represented as \( f(x) = \sqrt{x} \), is a specific type of radical function. It only returns non-negative results. This nature is important when considering inverses, particularly because an inverse should also reflect the domain and range adjustments.Key points to understand with square root functions:
- The Function's Domain: It is strictly the set of non-negative real numbers, \( x \geq 0 \). This is because you can't take the square root of a negative number without getting into imaginary numbers.
- Range: The output is also restricted to non-negative numbers. For \( f(x) = \sqrt{x+1} \), the domain starts from \( x = -1 \) (since \( x+1 \) needs to be non-negative), and the range is from \( 0 \) onward.
- Impact on Inversions: Since the output and input are both non-negative, any inverse functions particularly need to adhere to this rule, limiting \( x \) to values where the square root is defined.
